## Fourier Analysis and Number theory I: The self-dual function e^{-\pi x^2}

During a lecture at UCLA, Serge Lang asked what is the most important function in mathematics. This question is quite personal and every person certainly has his own opinion. In fact, a professor spoke out loud that the constant function is the most important function.

Serge Lang suggested ${f(x) = e^{-\pi x^2}}$. The function is self-dual (i.e. its Fourier transform is itself, see below) and the property is essential to the functional equation of the Riemann zeta function.

For ${f \in L^1({\mathbb R})}$, the Fourier transformof ${f}$ is a function ${\hat{f}}$ on ${{\mathbb R}}$ defined by

$\displaystyle \widehat{f}(t) = \int_{{\mathbb R}} f(x)e^{-2\pi i x t} dx.$

Since ${f \in L^1({\mathbb R})}$, the above integral is absolutely convergent.

Theorem 1 The function ${f(x) = e^{-\pi x^2}}$ is self-dual, i.e., ${\hat{f}(t) = f(t)}$.

Proof 1.We will use Cauchy’s theorem from the theory of complex analysis. For another proof without the theory of complex analysis, see proof 2 below.

$\displaystyle \widehat{f}(t) = \int_{-\infty}^\infty e^{-\pi x^2-2\pi i x t} dx = \int_{-\infty}^\infty e^{-\pi (x+i t)^2} e^{-\pi t^2} dx = e^{-\pi t^2} \int_{\Im z= t} e^{-\pi z^2} dz.$

We will show that the integral on the right hand side is ${1}$ and completes the proof. Let ${\Gamma_T}$ be the contour of the rectangle with vertices ${-T}$, ${T}$, ${T+it}$, ${-T+it}$. By Cauchy’s theorem

$\displaystyle 0 = \int_{\Gamma_T} e^{-\pi z^2} dz$

$\displaystyle =\int_{-T}^T e^{-\pi x^2} dx + \int_0^t e^{-\pi (T+yi)^2} i dy - \int_{-T}^T e^{-\pi (x+it)^2} dx - \int_0^t e^{-\pi (-T+iy)^2} idy.$

Let ${T\rightarrow \infty}$, the second term and the fourth term ${\rightarrow 0}$. Thus

$\displaystyle \int_{\Im z = t} e^{-\pi z^2} dz = \int_{{\mathbb R}} e^{-\pi x^2} dx = 1.$

The last equality is a standard application of double integration in polar coordinates. ${\square}$
Before we proceed to the second proof, we need the following lemma.

Lemma 2 Suppose ${f \in L^1}$.
(i)If ${f}$ has a continuous derivative and ${f' \in L^1({\mathbb R})}$, then

$\displaystyle \widehat{f'}(t) = 2\pi i t \widehat{f}(t).$

(ii) Let ${(Mf)(x) = xf(x)}$. Suppose ${(Mf)(x)}$ is integrable, then

$\displaystyle \widehat{Mf}(t) = -\frac 1{2\pi i} \frac{d}{dt} \widehat{f}(t).$

Proof. Suppose ${f' \in L^1({\mathbb R})}$

$\displaystyle \widehat{f'}(t) = \int_{-\infty}^\infty f'(x) e^{-2\pi i x t} dx$

$\displaystyle = \left. f(x) e^{-2\pi i x t} \right|_{-\infty}^{\infty} + 2 \pi i t \int_{-\infty}^\infty f(x) e^{-2\pi i x t} dx = 2 \pi i t \hat{f}(t).$

In the last step we use the fact that ${f \in L^1({\mathbb R})}$ and hence ${f(\pm \infty) = 0}$. This proves (i). Next Suppose ${xf(x)}$ is integrable. Then ${\frac{d}{dt} f(x) e^{-2\pi i x t} = -2\pi i x f(x) e^{-2\pi i x t}}$ is also integrable. Hence

$\displaystyle \frac{d}{dt} \widehat{f}(t) = \frac{d}{dt} \int_{\mathbb R} f(x) e^{-2\pi i x t} dx$

$\displaystyle = -2\pi i \int_{\mathbb R} x f(x) e^{-2\pi i x t} dx = -2\pi i \widehat{Mf}(t).$

Thus (ii) follows. ${\square}$

Proof 2. Let ${f(x) = e^{-\pi x^2}}$. Simple calculation shows that

$\displaystyle f'(x) + 2 \pi x f(x) = 0.$

Applying the Fourier transform to the above equation, by the above lemme we obtain

$\displaystyle 2 \pi i t \widehat{f}(t) + i \frac{d}{dt} \widehat{f}(t) = 0.$

Solving the ordinary differential equation, we obtain

$\displaystyle \widehat{f}(t) = C e^{-\pi t^2},$

where ${C}$ is a constant. To find ${C}$, we set ${t=0}$

$\displaystyle C = \widehat{f}(0) = \int_{{\mathbb R}} e^{-\pi x^2} dx = 1.$

${\square}$

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